Shortest Leg (SL1)

In this subsection, the modeling of Shortest Leg corresponding to strategy SL1 is addressed. The following assumptions in addition to section 4.1.1 apply here.

• The execution order is determined to be SSRR.

• The second storage position is an open location within the no-cost zone between SP1 and RP1. If no such position is available, the cycle is performed randomly according to the underlying model.

3 We assume 5·tmastexactly like for the random execution. It may by possible that t0consists of 4 · tmast, see subsection 4.2.3

• The dimension of each storage lane has the same proportion as the rack.

Table 5.8 summarizes all necessary components for the model.

Underlying model SSRR model: P (SSRR; SRSR) = (1,0) Changed components none

New components P (SL)

Table 5.8: Components for the derivation of the SL1 model

The aim of the shortest leg idea is to save one travel between distance, which can be realized if there is an empty position within the no-cost zone between SP1 and RP1. The travel time components of the cycle in compar-ison to a random executed cycle are shown in Figure 5.12.

x

Figure 5.12: Example of a SL1 cycle with the path-depending components of the cycle

No new analytical expressions are needed for the path-depending travel movements. Instead, we need to define the probability that there is an empty position within the no-cost zone, P(SL). Therefore, the size of the no-cost zone and the storage allocation within the zone are required.

Consider any mean travel between two randomly chosen positions, (x₁, y₁) and (x₂, y₂), within a normalized, dimensionless rack as shown in Figure 5.13. From section 3.2.1, we know the expected distance between them is

7/15. One of the two distances, in x- or y-direction, is the greater one and therefore determines the travel time. Without loss of generality, let |x1− x₂| ≥ |y1− y2|, then |x1− x2| is the normalized travel time. In the direction with the smaller distance, there is a range for additional movement of the S/R machine. This range is |x1− x2| − |y1− y2| as illustrated in Figure 5.13.

Within this range, the S/R machine can move to a greater area without loss of time. Figure 5.13 illustrates that from the angleα, that is formed by the horizontal axis and the connection between the two positions, the range is determined. (Note that for |x1− x2| ≤ |y1− y2|, the angle is formed by the vertical and the connecting line).

y

Range for additional movement of the S/R machine in y-direction during travel between

Figure 5.13: Travel between in a normalized, dimensionless rectangle showing the range in the shorter direction and the angleα.

To derive the mean size of the no-cost zone, the average value ofα is needed withα ∈ [0,45^◦]. With the mean value theorem of integral calculations, the expected distance in the smaller direction and thus the average value of α can be calculated. The result is α = 23.81049^◦(Brunk 2016). Knowing the average value ofα, the range for additional movement is₁₅⁷ −₁₅⁷ tan(α).

This is used to define the exact dimensions of the mean no-cost zone as shown in Figure 5.14.

To calculate the surface area of the no-cost zone (A_no−cost) its side lengths are needed (within the normalized (1x1)-rack). The line between SP1 and P2 is the hypotenuse of the right, isosceles triangle with leg length of 0.13037. Using the theorem of Pythagoras, the length is determined as 0.184371. The line between P2 and RP1 is the hypotenuse of the right, isosceles triangle with length of 0.336297. Using the theorem of Pythago-ras, the length is determined as 0.475595.

𝛼 = 23.81049°

15− 0.205927 = 0.26073

0.13037

Bc. moving forth and back 0.205927: 2 = 0.13047

Figure 5.14: Dimensions of the mean no-cost zone

The mean size of the no-cost zone can be calculated according to its di-mension, which is:

A_no−cost= 0.184371 · 0.475595 = 0.087686 (5.16) Given the total number of storage lanes of the rack (l ), the mean number of storage lanes within the no-cost zone can be determined:

N l_no−cost= Ano−cost· l = 0.087686 · l (5.17)

Knowing the mean total number of positions in the no-cost zone as well as the mean state probabilities of the storage lanes, we can determine the ex-pected number of available storage positions within the no-cost zone and thus the probability of finding at least one position that is available for stor-age. We can use the complementary probability, i.e., that no position is available within the no-cost zone. With P (F ) from the SSRR model (equa-tion 4.54), the probability that all posi(equa-tions within the no-cost zone are fully occupied is P (F )^{N lno−cost}. Consequently, the probability that performing the shortest leg cycle is possible is:

P (SL) = 1 − P(F )^{N lno−cost} (5.18)

If the shortest leg cycle is performed, E (QC^3S)^N applies for the path-depending travel time. Dwell times and acceleration parts remain un-changed as the number of stops is not reduced. Using equation 4.59 in case no shortest leg cycle is possible, the mean travel time of the main routing policy Shortest Leg is given by:

E (QC_{d d})SL= t0+5 2· (v_x

ax+v_y ay

)

+ (1 − P (SL)) · E(QC )^N· L

v_x + P (SL) · E(QC^3S)^N· L v_x + 1 · PRear r ang e· E(RC ) + 1 · PRear r ang e· tTang o

+ 4 · E(tLH D^S ) + 4 · E(t_{LH D}^R ) − 2 · tLH D, f

(5.19)

Note that equations 4.40, 4.42, 4.55, 4.56, 4.57 and 4.58 apply here.

It is important to mention that the derivation of the area of the no-cost zone is determined based on the continuous rack model. Transferring the result to a discrete rack with a finite number of storage lanes (l ), causes inaccura-cies due to discretization and rounding errors.